package leetcode.editor.cn;

//给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。
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// 示例 1： 
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//输入：head = [1,2,3,4,5]
//输出：[5,4,3,2,1]
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// 示例 2： 
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//输入：head = [1,2]
//输出：[2,1]
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// 示例 3： 
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//输入：head = []
//输出：[]
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// 提示： 
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// 链表中节点的数目范围是 [0, 5000] 
// -5000 <= Node.val <= 5000 
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// 进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？ 
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// Related Topics 递归 链表 👍 2670 👎 0

//Java：【206】- 反转链表
public class LeetCode_206_ReverseLinkedList{
    public static void main(String[] args) {
        Solution solution = new LeetCode_206_ReverseLinkedList().new Solution();
        // TO TEST
    }
    
    //leetcode submit region begin(Prohibit modification and deletion)
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    /* 迭代 */
    public ListNode reverseList(ListNode head) {
        ListNode prev =null;
        ListNode curr =head;
        while (curr!=null){
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev; //最后的prev就是tail, curr是tail.next=null
    }

}
//leetcode submit region end(Prohibit modification and deletion)
public class ListNode {
    int val;
    ListNode next;
    ListNode() {}
    ListNode(int val) { this.val = val; }
    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}
    /* 递归 */
    public ListNode reverseList(ListNode head) {
        if(head==null || head.next ==null) return head;
        ListNode newHead = reverseList(head.next);
        head.next.next = head;
        head.next =null;
        return newHead;
    }
}